candle flame or boiling water

[PalmRunner]

Hello guys.

I have a simple question: Let’s consider two heat sources for a LTD Stirling - a cup of boiling water and a candle flame. Assuming that the engine would survive the flame temperature, I just wonder which heat source would spin the engine to higher rpm?

In other words,
If the bottom plate area is ~ 190cm^2 (1mm thick alu disc)
What would be the average temperature of the bottom plate if:
500’C are applied on 1cm^2
80’C are applied on 50cm^2

Don’t know how to calculate this, but my intuition tells me that a cup of boiling water would heat the bottom plate to a higher average temperature. What do you think?

Totyo

 

[gilessim]

You have to consider that an LTD requires only a temperature difference of about 5° c-10°c between the top and bottom plates to run, more heat doesn't necessarily mean more speed, that's why they are called low temp. engines, in theory ,if the bottom plate was heated to 500°c then the top plate would still have to be 490-495°c for it to run, thus defeating the object!, for more speed you need to have more mass (i.e. a heatsink) on the top plate to dissipate the heat more quickly, but saying that doesn't necessarily mean that more speed = more power!

I'm intrigued as to why you are after more speed, do you have a practical application in mind?

Giles

 

[PalmRunner]

Yes, you are right it is not good idea to heat a LTD with a flame, but I am asking in principle…
The point is that you apply high temperature, but in a small area - maybe 1cm^2 and the rest is exposed to ambient temperature, while if using a cup of hot water you apply less heat, but on much larger area. Forget, for speed and power, don’t have anything special behind. The question is only: In which case the bottom plate will be heated to a higher average temperature?

Totyo

 

[mklotz]

Heat flow and dissipation is a very complicated subject theoretically - lots of partial differential equations and obscure constants to determine. I think an experiment is in order to even come close to answering your question.

Some thoughts on the subject...

There are VOMs on the market now that have thermocouple inputs. That or a contactless electronic thermometer would go a long way towards answering your question. Both are otherwise handy tools to have in the shop.

If you have a plastic foam displacer (many LTDs do), you don't want to get the hot plate too hot, lest you melt the displacer! If you do the experiment I suggested, dismantle the engine, remove the displacer and heat the empty chamber to see which option creates the higher temperature on the hot plate.

The fact that an LTD will run on a 5 degC differential doesn't mean that you need to maintain a 5 degC differential - only that you have at least that much differential. A greater differential will improve the efficiency of the engine, which is given by the Carnot efficiency,

efficiency = (Th - Tc)/Th = 1 - Tc/Th

where Th and Tc are the hot and cold temperatures respectively, expressed in an absolute (Kelvin) scale.

 

[BobWarfield]

Marv's right. The issue is strictly the non-metal materials in an LTD can melt or be damaged by too high a temperature. Otherwise, the bigger the differential, the more energy is available to power the engine.

I have one of those IR thermometers. Didn't cost much on eBay. I use it to check bearing temps on machinery that's running. It's a helpful way to know if you have too much or too little preload (or if you're getting ready to lose a bearing!).

Cheers,

BW

 

[PalmRunner]

Guys, maybe the subject is misleading …
My dilemma is NOT whether I can use a candle as heat source for my LTD. The foam displacer would certainly melt in the center. My question was purely theoretical - forget for the LTD, consider only a simple alu disc heated either by a candle or by a cup of boiling water. In which case the alu disc will get heated to a higher average temperature?

Marv suggested to perform an experiment, since “the heat flow and dissipation is a very complicated subject”. I didn’t know that this task is so complex. Do you know for some calculators available online?

Bob, can you suggest me a good IR thermometer. What is the resolution of these devices and in what temperature range they are working?

Totyo

 

[mklotz]

The temperature rise is related to the heat supplied by the equation

Q = cm * m * deltaT

where:

Q = heat
cm = specific heat of material
m = mass of material
deltaT = change in temperaure

In a perfect world, if one could measure or calculate Q for the two scenarios you propose, one could compute the resulting changes in temperature and answer your question.

But this isn't a perfect world. While you're heating the bottom of the plate, the top is radiating heat away (so is the bottom and edge but we won't talk about that complication) via both radiation and convection - both complex phenomena. In a practical sense, it's almost impossible to accurately calculate just how much of the heat applied to the plate actually ends up changing the temperature of the plate*. To further complicate the matter, the heat transfer mechanisms for an open flame and water vapor (steam) are probably wildly different. Then one has to ask, does the ability of aluminum to absorb heat change with temperature? The questions and uncertainites go on and on. Heat transfer is one of the messiest problems of macro physics. Our inability to model it well partially accounts for the fact that most long-term weather models are worthless.

Perhaps somewhere someone sitting at the control console of a Cray supercomputer could come up with a meaningful estimate of Q supplied but in a home workshop, the only practical solution is going to be an experiment - and even there you're going to have to go to considerable lengths to control the experimental parameters in order to get meaningful results.

Just because the question is easy to ask doesn't mean it's easy to answer.

---
*In the laboratory, it's more common to use that equation to estimate the heat supplied by measuring the (more easily measured) change in temperature. Measuring the caloric content of fuels by burning them to heat water and noting the temperature increase of the water is a common procedure.

 

[PalmRunner]

Thank you Marv for spending time answering silly questions. You are absolutely right, the experiment would be the best way. The theory is obviously quite complex.

Totyo

 

[mklotz]

Bitteschoen.

Your questions are by no means silly. Asking such questions in a forum like this one serves a useful purpose (even if we can't answer them). The questions highlight the complexity that working engineers face daily while designing complex mechanisms such as heat engines.

Physics expands its horizons by asking seemingly simple questions. In the early days of thermodynamics, there were two competing answers to the simple question, "What is heat?" Some thought it was a physical substance that was passed from object to object. Others argued that it was nothing more than the state of agitation of the atoms of the material. The ingenious experimentation that evolved to resolve this debate led to the development of both the thermometer and the principle of calorimetry that I alluded to in my footnote.

All of relativity stems from Einstein asking himself the innocuous question, "What would the world look like if I could ride on a beam of light?"

There are many more examples but in physics, as in model engineering, there are no silly questions.

 

[bigsteve]

Wow! 2 science questions in a day. There are some factors to determine: How well insulated is your cup of boiling water, and is it on a hot plate of some sort to keep it boiling?

We know that the candle is not going to be well insulated, otherwise, it would choke of air and go out. It will also depend on the type of candlewax (paraffin, eicosane, beeswax, etc.). The heat of combustion of paraffin is 42 kJ/g, so a 4 ounce candle that burns completely in 6 hours (I'm just guessing on these) will release 902 BTU/hr.

I'm going to assume your theoretical sterling engine can dissipate as much energy as it will take in - not likely in real life, but we're not talking about a real engine, so I say it's okay. That means that we can ignore the efficiency of heat absorption by the engine and just look at the energy released by the candle/water to room temperature air.

If you have the candle 2 inches below the base, the area of a sphere 2" in radius is 324 cm^2, so your 50 cm^2 base will absorb (assuming it is matte black) 50/324ths of the heat put out by the candle - 2.3 BTU/minute.

On to the boiling water. I'm assuming that you have a typical coffee cup and it is not on a hot plate. I will assume the initial cooling to be 1°C per 2 minute initially (at 80°C -again, these are all estimates), the water weighs 250 grams, the specific heat of water is 4.18 J/g*K, so the cup will release 4.18 J/g*K * 250 g * 1K in 2 minute, or 1.98 BTU/minute (sorry for switching between metric and imperial). I am also assuming that 10% of the heat loss is through the sides of the cup, so 1.78 BTU/minute will be absorbed by the engine.

These numbers are pretty close, and considering I guessed for some of the values, I would say it's probably a toss up for original power output, but over the long run (as the cup cools off but the candle continues to burn), the candle will put out more power.